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If the density of the wood is 3 pounds per cubic foot and if the weight of the solid is
360 pounds, what is the width, w, in feet, of the solid?
(A) 5.0
(B) 2.5
(C) 2.4
(D) 1.5

The area of the triangular face of the solid is 48 if that helps

If The Density Of The Wood Is 3 Pounds Per Cubic Foot And If The Weight Of The Solid Is 360 Pounds What Is The Width W In Feet Of The Solid A 50 B 25 C 24 D 15 class=

Sagot :

Abidemiokinidn

The width in feet of the given solid is 1.136

Density is defined as the ratio of the mass per unit volume

Density = Mass/Volume

Given

Weight = 360 pounds

Density = 3 pounds per cubic foot

Get the mass;

Since W = mg

m = W/g

m = 360/2.2

m =163.64  lbm/kg

Get the volume

From the formula;

Volume = Mass/Density

Volume = 163.64 /3

Volume = 54.55 ft³

Next is to get the width

Volume of the triangular prism = Base Area * Width

Base Area = 48ft²

54.55 = 48w

w = 54.55/48

w = 1.136

Hence the width of the solid in feet is 1.136ft

Learn more on how to calculate density here https://brainly.com/question/17887628

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